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T1 and E1 bit rate and bandwidth; DPCM encoder

Short answer

T1 bit rate is 1.544 Mb/s, min BW 772 kHz; E1 bit rate is 2.048 Mb/s, min BW 1.024 MHz. DPCM encodes prediction error, needing fewer bits than PCM.

What are the signalling (bit) rate and bandwidth requirement for the T1 and E1 digital carrier systems? Explain briefly about Differential Pulse Code Modulation (DPCM) encoder.

Bit rate and bandwidth of T1 and E1

Both systems sample speech at fs=8f_s = 8 kHz with 8 bits per sample.

T1: 24 channels + 1 framing bit:

Rb=(24×8+1)×8000=193×8000=1.544 Mb/sR_b = (24 \times 8 + 1) \times 8000 = 193 \times 8000 = 1.544\ \text{Mb/s}

E1: 32 slots (30 voice + 2 overhead):

Rb=32×8×8000=256×8000=2.048 Mb/sR_b = 32 \times 8 \times 8000 = 256 \times 8000 = 2.048\ \text{Mb/s}

Bandwidth. The minimum (Nyquist) bandwidth for binary transmission is B=Rb/2B = R_b/2. With practical NRZ/AMI pulses, the first-null bandwidth is about RbR_b.

SystemBit rateMin. bandwidth Rb/2R_b/2First-null BW ≈ RbR_b
T11.544 Mb/s772 kHz1.544 MHz
E12.048 Mb/s1.024 MHz2.048 MHz

DPCM encoder

Differential PCM uses the fact that speech samples next to each other are strongly correlated. It does not quantise each sample directly. It quantises the difference between the sample and a prediction made from earlier samples. The difference is smaller than the sample, so fewer bits are needed for the same quality.

 m[n] ->(+)--e[n]-->[Quantizer]--eq[n]--+-->[Encoder]-> bits
        ^-                              |
        |                               v
        |                   m^[n] ---> (+)
        |                     ^         | mq[n]
        +-------- m^[n] --[Predictor]<--+
  • Prediction error: e[n]=m[n]−m^[n]e[n] = m[n] - \hat m[n].
  • Quantised error: eq[n]=e[n]+q[n]e_q[n] = e[n] + q[n].
  • Predictor input: mq[n]=m^[n]+eq[n]=m[n]+q[n]m_q[n] = \hat m[n] + e_q[n] = m[n] + q[n].

The predictor (a linear filter on past values of mqm_q) works on the quantised samples. The receiver has the same predictor, so encoder and decoder stay in step and quantisation errors do not build up.

The SNR improves by the prediction gain Gp=σm2/σe2G_p = \sigma_m^2 / \sigma_e^2. DPCM saves about 1–2 bits per sample compared with PCM, e.g. 48–56 kb/s instead of 64 kb/s for speech. Adaptive DPCM (ADPCM) goes further, to 32 kb/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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